Question on Ionic Equilibrium
A student dissolved a certain amount of a weak acid (HA) in some water. She then titrated this solution against NaOH of some unknown concentration. When she had added 5mL of NaOH the pH of the solution was found to be 5. It took a total of 20mL of NaOH to completely neutralize the acid solution she had prepared. What is the acid dissociation constant (\(K_a\)) of the weak acid, HA?
Answer
let us assume the total moles of the acid in solution as “x”.
20mL of NaOH is required to completely neutralize the acid solution. This means that 5mL of NaOH will be able to neutralize \(\frac{1}{4}\times x\) moles of the acid taken and produce \(\frac{1}{4} \times x\) moles of the salt of this acid with a strong base. The resulting solution will be an acidic buffer.
pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[pH = pK_a + log\dfrac{[A^-]}{[HA]}\]
We have \(\frac{3}{4}\times x\) moles of the acid (\([HA]\)) remaining and \(\frac{1}{4} \times x\) moles of the salt (\(A^-\)). The pH of the solution was 5 at this point of the titration.
So,
\[pH = pK_a + log\dfrac{[A^-]}{[HA]}\]
\[5 = pK_a + log\dfrac{\frac{1}{4}}{\frac{3}{4}}\]
\[5 = pK_a + log\dfrac{1}{3}\]
\[5 = pK_a - log3\]
\[5 = pK_a - 0.477\]
\[pKa = 5 + 0.477\]
\[pKa = 5.477\]
\[-log K_a = 5.477\]
\[K_a = 3.33 \times 10^{-6}\]
